Question

如果我有以下列表：

[＆＃39;主持人：www.perdu.com＆＃39;，＆＃39;连接：keep-alive＆＃39;，＆＃39;缓存控制： max-age = 0＆＃39;，＆＃39;接受： text / html的，应用/ XHTML + xml的，应用/ XML; Q = 0.9，图像/ WEBP， / 的; Q = 0.8＆＃39 ;, ＆＃39; User-Agent：Mozilla / 5.0（Macintosh; Intel Mac OS X 10_9_4） AppleWebKit / 537.36（KHTML，与Gecko一样）Chrome / 38.0.2125.111 Safari / 537.36＆＃39;，＆＃39; DNT：1＆＃39;，＆＃39;接受编码：gzip，deflate，sdch＆＃39;，＆＃39;接受 - 语言：en-US，en; q = 0.8，es; q = 0.6，fr; q = 0.4＆＃39;，＆＃39;＆＃39;]

我怎样才能得到一个元组列表tuple_list，如：

tuples_list = [
    ('Host', 'www.perdu.com'),
    ('Connection', 'keep-alive'),
     ....
]

Answer 1

您听说过list comprehension吗？

headers = ['Host: www.perdu.com', 'Connection: keep-alive', 'Cache-Control: max-age=0', 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8', 'User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36', 'DNT: 1', 'Accept-Encoding: gzip,deflate,sdch', 'Accept-Language: en-US,en;q=0.8,es;q=0.6,fr;q=0.4', '']
headers = [i.split(': ') for i in headers]

会回复你：

[['Host', 'www.perdu.com'], ['Connection', 'keep-alive'], ['Cache-Control', 'max-age=0'], ['Accept', 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8'], ['User-Agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36'], ['DNT', '1'], ['Accept-Encoding', 'gzip,deflate,sdch'], ['Accept-Language', 'en-US,en;q=0.8,es;q=0.6,fr;q=0.4'], ['']]

如果你这样做，你可以得到一个元组列表：

headers = ['Host: www.perdu.com', 'Connection: keep-alive', 'Cache-Control: max-age=0', 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8', 'User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36', 'DNT: 1', 'Accept-Encoding: gzip,deflate,sdch', 'Accept-Language: en-US,en;q=0.8,es;q=0.6,fr;q=0.4', '']
headers = [tuple(i.split(': ')) for i in headers]

这将返回您的期望：

[('Host', 'www.perdu.com'), ('Connection', 'keep-alive'), ('Cache-Control', 'max-age=0'), ('Accept', 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8'), ('User-Agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36'), ('DNT', '1'), ('Accept-Encoding', 'gzip,deflate,sdch'), ('Accept-Language', 'en-US,en;q=0.8,es;q=0.6,fr;q=0.4'), ('',)]

Answer 2

怎么样：

tuples_list = [tuple(item.split(": ")) for item in thelist]

这使用列表推导来迭代列表中的每个项目，将其拆分为:，并将每个结果数组转换为元组。

Answer 3

您可以在列表推导中使用split(':')来实现这一目标。

>>> tuples_list = [tuple(i.split(':')) for i in l]
>>> tuples_list
[('Host', ' www.perdu.com'),
 ('Connection', ' keep-alive'),
 ('Cache-Control', ' max-age=0'),
 ('Accept', ' text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8'),
 ('User-Agent', ' Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36'),
 ('DNT', ' 1'),
 ('Accept-Encoding', ' gzip,deflate,sdch'),
 ('Accept-Language', ' en-US,en;q=0.8,es;q=0.6,fr;q=0.4')]

将带有冒号分隔项的列表拆分为元组列表

3 个答案: